Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        Segment[] segs;
        double a;
        double b;
        double sy;
        int N;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            sy = in.nextInt();
            a = in.nextInt();
            b = in.nextInt();
            N = in.nextInt();
            segs = new Segment[N];
            FenwickTree fen = new FenwickTree(N);
            for (int i = 0; i < N; i++) {
                segs[i] = new Segment(in.nextInt(), in.nextInt());
                fen.add(i, (int) (segs[i].ri - segs[i].li));
            }
            int q = in.nextInt();
            while (q-- > 0) {
                double xi = in.nextInt();
                double yi = in.nextInt();
                int left = findLeftEdge(xi, yi);
                int right = findRightEdge(xi, yi);
                double res = 0;
                if (left != N) {
                    res += getOverlap(segs[left], xi, yi);
                }
                if (right != -1 && right != left) {
                    res += getOverlap(segs[right], xi, yi);
                }
                res += (yi - sy) / yi * fen.sum(left + 1, right - 1);
                if (equals(res, 3061225.025)) {
                    out.println(left + " " + right + " " + xi + " " + yi);
                }
                out.println(res);
            }
        }

        boolean equals(double x, double y) {
            return (Math.abs(x - y) <= 0.000001);
        }

        double getOverlap(Segment s, double xi, double yi) {
            double right = ((xi - s.ri) * sy + yi * s.ri) / yi;
            double left = ((xi - s.li) * sy + yi * s.li) / yi;
            left = Math.max(a, left);
            right = Math.min(b, right);
            return Math.max(0, right - left);
        }

        int findLeftEdge(double xi, double yi) {
            int low = 0;
            int high = N;
            while (low < high) {
                int mid = (low + high) / 2;
                if (((xi - segs[mid].ri) * (sy / yi) + segs[mid].ri) >= a) {
                    high = mid;
                } else {
                    low = mid + 1;
                }
            }
            return high;
        }

        int findRightEdge(double xi, double yi) {
            int low = -1;
            int high = N - 1;
            while (low < high) {
                int mid = (low + high + 1) / 2;
                if (((xi - segs[mid].li) * (sy / yi) + segs[mid].li <= b)) {
                    low = mid;
                } else {
                    high = mid - 1;
                }
            }
            return low;
        }

        class Segment {
            long li;
            long ri;

            Segment(int li, int ri) {
                this.li = li;
                this.ri = ri;
            }

        }

    }

    static class FenwickTree {
        public int[] BIT;
        public int size = 0;

        public FenwickTree(int N) {
            BIT = new int[N];
            size = N;
        }

        public FenwickTree(int[] arr) {
            size = arr.length;
            BIT = new int[arr.length];
            for (int i = 0; i < size; i++) {
                add(i, arr[i]);
            }
        }

        public void add(int id, int add) {
            for (int i = id; i < size; i |= i + 1) {
                BIT[i] += add;
            }
        }

        public int sum(int l, int r) {
            if (r < l) {
                return 0;
            }
            return sum(r) - sum(l - 1);
        }

        public int sum(int r) {
            if (r < 0 || r >= size) {
                return 0;
            }
            int res = 0;
            for (int i = r; i >= 0; i = ((i) & (i + 1)) - 1) {
                res += BIT[i];
            }
            return res;
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}